The value of [tex]$\sin \frac{5\pi }{4}$[/tex] and [tex]$\cos \frac{5\pi }{4}$[/tex] is [tex]$-\frac{\sqrt{2}}{2}$[/tex] and [tex]$\tan \frac{5\pi }{4}$[/tex] is [tex]$1$[/tex] .
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The unit circle has a radius of one and is centered at the origin. The unit circle's equation is
[tex]x^2+y^2=1[/tex]
We have to find the value of [tex]$\sin \frac{5\pi }{4}$[/tex], [tex]$\cos \frac{5\pi }{4}$[/tex] and [tex]$\tan \frac{5\pi }{4}$[/tex].
Write [tex]$\sin \left( \frac{5\pi }{4} \right)$[/tex]as [tex]$\sin \left( \pi +\frac{\pi }{4} \right)$[/tex].
[tex]$=\sin \left( \pi +\frac{\pi }{4} \right)$[/tex]
Use the summation identity,
[tex]$\sin \left( x+y \right)=\sin \left( x \right)\cdot \cos \left( y \right)+\cos \left( x \right)\cdot \sin \left( y \right)$[/tex]
[tex]$=\sin \left( \pi \right)\cdot \cos \left( \frac{\pi }{4} \right)+\cos \left( \pi \right)\cdot \sin \left( \frac{\pi }{4} \right)$[/tex]
Use the following trivial identity,
[tex]$\cos \left( \pi \right)=\left( -1 \right)$[/tex]
[tex]$\sin \left( \pi \right)=0$[/tex]
[tex]$\cos \left( \frac{\pi }{4} \right)=\frac{\sqrt{2}}{2}$[/tex]
And
[tex]$\sin \left( \frac{\pi }{4} \right)=\frac{\sqrt{2}}{2}$[/tex]
So,
[tex]$=0\cdot \frac{\sqrt{2}}{2}+\left( -1 \right)\cdot \frac{\sqrt{2}}{2}$[/tex]
[tex]$\therefore \sin \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}$[/tex]
The value for [tex]$\cos \frac{5\pi }{4}$[/tex] is same as [tex]$\sin \frac{5\pi }{4}$[/tex].
Solve for [tex]$\tan \frac{5\pi }{4}$[/tex].
Rewrite
[tex]$\tan \frac{5\pi }{4}=\tan \frac{\pi }{4}$[/tex]
Use the following trivial identity,
[tex]$\tan \left( \frac{\pi }{4} \right)=1$[/tex]
[tex]$\therefore \tan \left( \frac{5\pi }{4} \right)=1$[/tex]
Hence, the value is,
[tex]$\sin \left( \frac{5\pi }{4} \right)=-\frac{\sqrt{2}}{2}$[/tex],
[tex]$\cos \left( \frac{5\pi }{4} \right)=-\frac{\sqrt{2}}{2}$[/tex],
and
[tex]$\tan \left( \frac{5\pi }{4} \right)=1$[/tex]
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